Problem: You have found the following ages (in years) of all 6 turtles at your local zoo: $ 76,\enspace 35,\enspace 38,\enspace 71,\enspace 86,\enspace 66$ What is the average age of the turtles at your zoo? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we have data for all 6 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{76 + 35 + 38 + 71 + 86 + 66}{{6}} = {62\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $76$ years $14$ years $196$ years $^2$ $35$ years $-27$ years $729$ years $^2$ $38$ years $-24$ years $576$ years $^2$ $71$ years $9$ years $81$ years $^2$ $86$ years $24$ years $576$ years $^2$ $66$ years $4$ years $16$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{196} + {729} + {576} + {81} + {576} + {16}} {{6}} $ $ {\sigma^2} = \dfrac{{2174}}{{6}} = {362.33\text{ years}^2} $ The average turtle at the zoo is 62 years old. The population variance is 362.33 years $^2$.